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kashbmaryd
Group Title
Hello everyone. It's been awhile since I've been up here. I have one about a polynomial / another polynomial. Here it is:
(x^2y^2+2yzz^2)/(x+yz)
Any thoughts?
 one year ago
 one year ago
kashbmaryd Group Title
Hello everyone. It's been awhile since I've been up here. I have one about a polynomial / another polynomial. Here it is: (x^2y^2+2yzz^2)/(x+yz) Any thoughts?
 one year ago
 one year ago

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whpalmer4 Group TitleBest ResponseYou've already chosen the best response.1
You do long division on polynomials just like on numbers. What is the biggest thing you could multiply (x+yz) by and have the result be < or = to the numerator? After you subtract the result from the numerator, what is left? Is there another value you could multiply (x+yz) by that would be less than or equal to that? If so, that's another term in your answer, otherwise, what's left is the remainder, if it isn't 0.
 one year ago

kashbmaryd Group TitleBest ResponseYou've already chosen the best response.0
It's a long division problem. The book answer is xy+z. But how do you get there doing the long division?
 one year ago

kashbmaryd Group TitleBest ResponseYou've already chosen the best response.0
It would be pretty easy if it was just a poly / a monomial. But this thing was tricky to me.
 one year ago

kashbmaryd Group TitleBest ResponseYou've already chosen the best response.0
Also I don't see how you could subtract the give denominator from the given numerator when there aren't any like terms on one versus the other.
 one year ago

kashbmaryd Group TitleBest ResponseYou've already chosen the best response.0
A little easier said that done huh?
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.1
You gotta have faith :) \[(x^2  y^2 + 2yz  z^2) / (x + y  z)\] Okay, by looking it, we know our answer has to have an x, a y, and a z, right? That's the only way we'll get x^2, y^2 and z^2 terms. We'll do this as a "speculative division" — we'll speculate about the answer, and then see if works. What we do will also bear some similarity to solving multiple equations with combination. Let's make a little table of what {x, y, z} * (x+yz) gives us: \[x(x+yz) = x^2 + xy  xz\]\[y(x+yz) = y^2 + xy  yz\]\[z(x+yz) = z^2 + xz + yz\] Those are our building blocks. How do we combine them to get our original numerator? Well, we need an x in the quotient to produce the x^2 term. We need a y in the quotient to produce the y^2 term. We need a z in the quotient to produce the z^2 term. Let's just add up those equations and see if we get the right thing! \[x(x+yz)  y(x+yz) + z(x+yz)=\]\[x^2+xyxz  (y^2+xyyz) + (z^2+xz+yz)=\] \[x^2 + xy  xz y^2 xy (yz) z^2 +xz +yz=\]\[x^2y^2 z^2 + xy xy +xz xz +yz+yz \] and after the dust settles and we cross off all the terms that combine to 0, we have\[x^2y^2+2yz z^2\] which is our original numerator. Therefore, the quotient we "guessed", (x  y + z), is correct!
 one year ago

kashbmaryd Group TitleBest ResponseYou've already chosen the best response.0
It took me a few minutes. Let me look at what you have there. Thanks for the help too.
 one year ago

kashbmaryd Group TitleBest ResponseYou've already chosen the best response.0
I will have to test that theory on other problems, but thanks for that. I really appreciate it. However, do you know how to do it using the long division?
 one year ago

kashbmaryd Group TitleBest ResponseYou've already chosen the best response.0
In other words, suppose I don't have a guess. Then what?
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.1
Well, it really *is* what you do when you do long division, I just organized the work a little differently.
 one year ago

kashbmaryd Group TitleBest ResponseYou've already chosen the best response.0
Oh ok. I got you.
 one year ago

kashbmaryd Group TitleBest ResponseYou've already chosen the best response.0
I will try that over the weekend on some other problems.
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.1
x  y + z x+yz x^2  y^2 + 2yz  z^2 x^2 xy xz  y^2 + 2yz  z^2  xy +xz y^2 +yz  xy  yz  z^2 +xz yz z^2 +xz 
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.1
I tend to make mistakes when doing lots of algebra, so if I can think a little first and simplify the pencilpushing, my accuracy goes up :)
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.1
Writing it out like a long division problem, it took me a couple of tries before I avoided making any mistakes. Doing it the "combination" way, I did it the first time, and it took less time, too!
 one year ago

kashbmaryd Group TitleBest ResponseYou've already chosen the best response.0
That was the first time I've seen it done that way. I am going to look that up on youtube later.
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.1
I agree it's pretty weird seeing all of these "other" terms popping up as you do the long division, but like I said, you just have to have faith that they will all go away in the end :)
 one year ago

kashbmaryd Group TitleBest ResponseYou've already chosen the best response.0
It kind of threw me for a loop, but I will have this figured out by tomorrow.
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.1
That's the attitude I like to hear! :) Too many people just say "oh, this is hard" or "I'm dumb" or whatever. Much of this stuff isn't really hard at all — it just takes the right explanation, and practice.
 one year ago

kashbmaryd Group TitleBest ResponseYou've already chosen the best response.0
Well, thanks again.
 one year ago

kashbmaryd Group TitleBest ResponseYou've already chosen the best response.0
I'm a fan now. Ttyl.
 one year ago

RA1dALLDAY Group TitleBest ResponseYou've already chosen the best response.0
Use PEMDAS!
 one year ago
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